Tan 20^o.Tan 40^o.Tan 80^o এর মান কত?

Explanation:

Another Explanation (5): bài toán: Tan 20^o.Tan 40^o.Tan 80^o = ? আমরা জানি, \( tan(3\theta) = \frac{3tan(\theta) - tan^3(\theta)}{1 - 3tan^2(\theta)} \) এখন, \( \theta = 20^\circ \) হলে, \( 3\theta = 60^\circ \) \( tan(60^\circ) = \frac{3tan(20^\circ) - tan^3(20^\circ)}{1 - 3tan^2(20^\circ)} \) \( \sqrt{3} = \frac{3tan(20^\circ) - tan^3(20^\circ)}{1 - 3tan^2(20^\circ)} \) উভয় পক্ষে \( (1 - 3tan^2(20^\circ)) \) গুণ করে পাই, \( \sqrt{3} (1 - 3tan^2(20^\circ)) = 3tan(20^\circ) - tan^3(20^\circ) \) --- (1) আমাদের প্রদত্ত রাশিমালা: \( tan(20^\circ) . tan(40^\circ) . tan(80^\circ) \) = \( tan(20^\circ) . tan(60^\circ - 20^\circ) . tan(60^\circ + 20^\circ) \) আমরা জানি, \( tan(A + B) = \frac{tanA + tanB}{1 - tanA.tanB} \) এবং \( tan(A - B) = \frac{tanA - tanB}{1 + tanA.tanB} \) সুতরাং, \( tan(60^\circ - 20^\circ) = \frac{tan60^\circ - tan20^\circ}{1 + tan60^\circ . tan20^\circ} = \frac{\sqrt{3} - tan20^\circ}{1 + \sqrt{3}tan20^\circ} \) এবং \( tan(60^\circ + 20^\circ) = \frac{tan60^\circ + tan20^\circ}{1 - tan60^\circ . tan20^\circ} = \frac{\sqrt{3} + tan20^\circ}{1 - \sqrt{3}tan20^\circ} \) এখন, \( tan(20^\circ) . tan(40^\circ) . tan(80^\circ) \) = \( tan(20^\circ) . \frac{\sqrt{3} - tan20^\circ}{1 + \sqrt{3}tan20^\circ} . \frac{\sqrt{3} + tan20^\circ}{1 - \sqrt{3}tan20^\circ} \) = \( tan(20^\circ) . \frac{(\sqrt{3} - tan20^\circ) . (\sqrt{3} + tan20^\circ)}{(1 + \sqrt{3}tan20^\circ) . (1 - \sqrt{3}tan20^\circ)} \) = \( tan(20^\circ) . \frac{3 - tan^2(20^\circ)}{1 - 3tan^2(20^\circ)} \) = \( \frac{3tan(20^\circ) - tan^3(20^\circ)}{1 - 3tan^2(20^\circ)} \) = \( tan(3 \times 20^\circ) = tan(60^\circ) = \sqrt{3} \) 🥳 সুতরাং, \( tan(20^\circ) . tan(40^\circ) . tan(80^\circ) = \sqrt{3} \)